2222 quantum physics 2007-81 2.1 an equation for the matter waves: the time- dependent schrődinger...

2222 Quantum Physics 2007-8 1

2.1 An equation for the matter waves: the time-dependent Schrődinger equation***

2 2

2 2 2

1 (2.1)

x v t

( , )x twave velocityv

Classical wave equation (in one dimension):

Can we use this to describe the matter waves in free space?

e.g. Transverse waves on a string:

x

( , ) wave displacement (in 1d)x t

Rae §2.1, B&J §3.1, B&M §5.1


An equation for the matter waves (2)2

2Try

t x

2 2

2i (2.2)

2t m x

Seem to need an equation that involves the first derivative in time, but the second derivative in space

(for matter waves in free space)

( , ) is "wave function" associated with matter wavex t


An equation for the matter waves (3)2

( , )2

pE V x t

m

2 2

2i ( , ) (2.3)

2V x t

t m x

For particle with potential energy V(x,t), need to modify the relationship between energy and momentum:

Suggests corresponding modification to Schrődinger equation:

Time-dependent Schrődinger equation

Total energy = kinetic energy + potential energy

Schrődinger


The Schrődinger equation: notes

•This was a plausibility argument, not a derivation. We believe the Schrődinger equation to be true not because of this argument, but because its predictions agree with experiment.

•There are limits to its validity. In this form it applies to

•A single particle, that is

•Non-relativistic (i.e. has non-zero rest mass and velocity very much below c)

•The Schrődinger equation is a partial differential equation in x and t (like classical wave equation)

•The Schrődinger equation contains the complex number i. Therefore its solutions are essentially complex (unlike classical waves, where the use of complex numbers is just a mathematical convenience)


The Hamiltonian operator2 2

2i ( , )

2V x t

t m x

2 2

2ˆ( , ) (2.4)

2

dV x t H

m dx

ˆi (2.5)Ht

Can think of the RHS of the Schrődinger equation as a differential operator that represents the energy of the particle.

This operator is called the Hamiltonian of the particle, and usually given the symbol

Hence there is an alternative (shorthand) form for time-dependent Schrődinger equation:

H

Kinetic energy operator

Potential energy operator


2.2 The significance of the wave function***

2 2

0( , ) ( , ) d

bb

xx a a

x t x x t x

2 * 2( , ) (2.6)x t x

Ψ is a complex quantity, so what can be its significance for the results of real physical measurements on a system?

Remember photons: number of photons per unit volume is proportional to the electromagnetic energy per unit volume, hence to square of electromagnetic field strength.

Postulate (Born interpretation): probability of finding particle in a small length δx at position x and time t is equal to

Note: |Ψ(x,t)|2 is real, so probability is also real, as required.

Total probability of finding particle between positions a and b is

a b

|Ψ|2

x

δx

Born

Rae §2.1, B&J §2.2, B&M §5.2


ExampleSuppose that at some instant of time a particle’s wavefunction is

( , ) 2 (for 0 0.909)

0 (otherwise)

x t x x

What is:

(a) The probability of finding the particle between x=0.5 and x=0.5001?

(b) The probability per unit length of finding the particle at x=0.6?

(c) The probability of finding the particle between x=0.0 and x=0.5?


NormalizationTotal probability for particle to be anywhere should be one (at any time):

Suppose we have a solution to the Schrődinger equation that is not normalized, Then we can

•Calculate the normalization integral

•Re-scale the wave function as

(This works because any solution to the S.E., multiplied by a constant, remains a solution, because the equation is linear and homogeneous)

Normalization condition2

( , ) d 1 (2.7)x t x

Alternatively: solution to Schrödinger equation contains an arbitrary constant, which can be fixed by imposing the condition (2.7)


Normalizing a wavefunction - example


2.3 Boundary conditions for the wavefunction

The wavefunction must:

1. Be a continuous and single-valued function of both x and t (in order that the probability density be uniquely defined)

2. Have a continuous first derivative (unless the potential goes to infinity)

3. Have a finite normalization integral.

Rae §2.3, B&J §3.1


2.4 Time-independent Schrődinger equation***

2 2

2i ( )

2V x

t m x

Suppose potential V(x,t) (and hence force on particle) is independent of time t:

LHS involves only variation of Ψ with t

RHS involves only variation of Ψ with x (i.e. Hamiltonian operator does not depend on t)

Look for a solution in which the time and space dependence of Ψ are separated:

( , ) ( ) ( ) (2.8)x t x T t

Substitute:

Rae §2.2, B&J §3.5, B&M §5.3


Time-independent Schrődinger equation (contd)

2 2

2

d( ) (2.11)

2 dV x E

m x

Solving the time equation:

The space equation becomes:

Time-independent Schrődinger equation

ˆ (2.12)H E or


Notes• In one space dimension, the time-independent Schrődinger equation is

an ordinary differential equation (not a partial differential equation)• The sign of i in the time evolution is determined by the choice of the

sign of i in the time-dependent Schrődinger equation• The time-independent Schrődinger equation can be thought of as an

eigenvalue equation for the Hamiltonian operator:Operator × function = number × function

(Compare Matrix × vector = number × vector) [See 2246]

• We will consistently use uppercase Ψ(x,t) for the full wavefunction (time-dependent Schrődinger equation), and lowercase ψ(x) for the spatial part of the wavefunction when time and space have been separated (time-independent Schrődinger equation)

• Probability distribution of particle is now independent of time (“stationary state”):

H E

2 2

2 2

2

( , ) exp( i / ) ( )

exp( i / ) ( )

( )

x t Et x

Et x

x

For a stationary state we can use either ψ(x) or Ψ(x,t) to compute probabilities; we will get the same result.


2.6 SE in three dimensions

H E

To apply the Schrődinger equation in the real (three-dimensional) world we keep the same basic structure:

ˆi Ht

BUT

Wavefunction and potential energy are now functions of three spatial coordinates:

Kinetic energy now involves three components of momentum

Interpretation of wavefunction:

Rae §3.1, B&J §3.1, B&M §5.1


Puzzle

( , ) exp[ ( )]x t i kx t

2 2 2 2 40E p c m c

The requirement that a plane wave

plus the energy-momentum relationship for free-non-relativistic particles

led us to the free-particle Schrődinger equation.

Can you use a similar argument to suggest an equation for free relativistic particles, with energy-momentum relationship:

2

2

pE

m


3.1 A Free Particle

2 2

2

d

2 dE

m x

Free particle: experiences no forces so potential energy independent of position (take as zero)

Time-independent Schrődinger equation:

Linear ODE with constant coefficients so try

exp( )x

Combine with time dependence to get full wave function:

General solution:


Notes

• Plane wave is a solution (just as well, since our plausibility argument for the Schrődinger equation was based on this being so)

• Note signs: – Sign of time term (-iωt) is fixed by sign adopted in time-

dependent Schrődinger Equation

– Sign of position term (±ikx) depends on propagation direction of wave

• There is no restriction on the allowed energies, so there is a continuum of states


3.2 Infinite Square Well

V V Consider a particle confined to a finite length –a<x<a by an infinitely high potential barrier

V(x)

x

-a a

0V

No solution in barrier region (particle would have infinite potential energy).

In well region:

Rae §2.4, B&J §4.5, B&M §5.4

Boundary conditions:

Continuity of ψ at x=a:

Continuity of ψ at x=-a:

Note discontinuity in dψ/dx allowable, since potential is infinite


Infinite square well (2)Add and subtract these conditions:

Even solution: ψ(x)=ψ(-x)

Odd solution: ψ(x)=-ψ(-x)

Energy:


Infinite well – normalization and notes

Notes on the solution:• Energy quantized to particular values (characteristic of bound-state problems in

quantum mechanics, where a particle is localized in a finite region of space.• Potential is even under reflection; stationary state wavefunctions may be even or

odd (we say they have even or odd parity)• Compare notation in 1B23 and in books:

– 1B23: well extended from x=0 to x=b– Rae and B&J: well extends from x=-a to x=+a (as here)– B&M: well extends from x=-a/2 to x=+a/2(with corresponding differences in wavefunction)

Normalization:


The infinite well and the Uncertainty Principle

Position uncertainty in well:

Momentum uncertainty in lowest state from classical argument (agrees with fully quantum mechanical result, as we will see in §4)

Compare with Uncertainty Principle:

Ground state close to minimum uncertanty


3.3 Finite square wellNow make the potential well more realistic by making the barriers a finite height V0

V(x)

x

-a a

V0

I II III

Region I: Region II: Region III:

0Assume 0 E V

Rae §2.4, B&J §4.6


Finite square well (2)

x a

Match value and derivative of wavefunction at region boundaries:

Add and subtract:

x a

Match ψ:

Match dψ/dx:


Finite square well (3)Divide equations:

Must be satisfied simultaneously:

Cannot be solved algebraically. Convenient form for graphical solution:


Graphical solution for finite well

-5-4-3-2-1012345

0 5 10

k

ktan(ka)

sqrt(k0^2-k^2)

kcot(ka)

-sqrt(k0^2-k^2)

k0=3, a=1


Notes• Penetration of particle into “forbidden” region where

V>E (particle cannot exist here classically)• Number of bound states depends on depth of

potential well, but there is always at least one (even) state

• Potential is even function, wavefunctions may be even or odd (we say they have even or odd parity)

• Limit as V0→∞:0 , circle becomes very large

Solutions at (as for infinite well)2

k

nka


Example: the quantum well

EsakiKroemer

Quantum well is a “sandwich” made of two different semiconductors in which the energy of the electrons is different, and whose atomic spacings are so similar that they can be grown together without an appreciable density of defects:

Electron potential energy

Position

Now used in many electronic devices (some transistors, diodes, solid-state lasers)

Material A (e.g. AlGaAs) Material B (e.g. GaAs)


3.4 Particle Flux

2 2exp[ ( )] d dA i kx t x A x

In order to analyse problems involving scattering of free particles, need to understand normalization of free-particle plane-wave solutions.

This problem is related to Uncertainty Principle:

Momentum is completely defined

Position completely undefined; single particle can be anywhere from -∞ to ∞, so probability of finding it in any finite region is zero

Conclude that if we try to normalize so that

2d 1x

will get A=0.

Rae §9.1; B&M §5.2, B&J §3.2


Particle Flux (2)More generally: what is rate of change of probability that a particle exists in some region (say, between x=a and x=b)?

xa b*

* *d db b

a a

x xt t t

Use time-dependent Schrődinger equation:


Particle Flux (3)Integrate by parts:

xa b

Interpretation:

Flux entering at x=a

Flux leaving at x=b

-

**

Particle flux at position

i( )

2

x

xm x x

Note: a wavefunction that is real carries no current

Note: for a stationary state can use either ψ(x) or Ψ(x,t)


Particle Flux (4)Sanity check: apply to free-particle plane wave.

# particles passing x per unit time = # particles per unit length × velocity

Makes sense:

Wavefunction describes a “beam” of particles.


3.5 Potential StepConsider a potential which rises suddenly at x=0:

x

V(x)

0

V0

Case 1: E<V0 (below step)

x<0 x>0

Boundary condition: particles only incident from left

Rae §9.1; B&J §4.3

Case 1


Potential Step (2)

Continuity of ψ at x=0:

dContinuity of at 0 :

dx

x

Solve for reflection and transmission:


Transmission and reflection coefficients


Potential Step (3)Case 2: E>V0 (above step)

Solution for x>0 is now

Matching conditions:

Transmission and reflection coefficients:


Summary of transmission through potential step

Notes:

•Some penetration of particles into forbidden region even for energies below step height (case 1, E<V0);

•No transmitted particle flux, 100% reflection (case 1, E<V0);

•Reflection probability does not fall to zero for energies above barrier (case 2, E>V0).

•Contrast classical expectations:

100% reflection for E<V0, with no penetration into barrier;

100% transmission for E>V0


3.6 Rectangular Potential BarrierRae §2.5; B&J §4.4; B&M §5.9

Now consider a potential barrier of finite thickness:

V(x)

x

a

I II III

0

V0

Boundary condition: particles only incident from left

Region I: Region II: Region III:

0Assume 0 E V


Rectangular Barrier (2)

0x

Match value and derivative of wavefunction at region boundaries:

x a

Match ψ:

Match dψ/dx:

Eliminate wavefunction in central region:


Rectangular Barrier (3)

Transmission and reflection coefficients:

For very thick or high barrier:

1a

Non-zero transmission (“tunnelling”) through classically forbidden barrier region:


Examples of tunnellingTunnelling occurs in many situations in physics and astronomy:

1. Nuclear fusion (in stars and fusion reactors)

3. Field emission of electrons from surfaces (e.g. in plasma displays)

V

2

0 nucleus

( )Barrier height ~ ~ MeV

4

thermal energies (~keV)

Ze

r

Internuclear distance x

Coulomb interaction (repulsive)

Strong nuclear force (attractive)

Incident particles

2. Alpha-decay

Distance x of α-particle from nucleus

V

Initial α-particle energy

Distance x of electron from surface

V

Work function W

VacuumMaterial


3.7 Simple Harmonic Oscillator

0

2 2 20

Force

Angular frequency =

1 1Potential energy ( ) 2 2

F kx

k

m

V x kx m x

Example: particle on a spring, Hooke’s law restoring force with spring constant k:

Mass m

x

Time-independent Schrődinger equation:

Problem: still a linear differential equation but coefficients are not constant.

Simplify: change variable to 1/ 2

0my x

Rae §2.6; B&M §5.5; B&J §4.7


Simple Harmonic Oscillator (2)

2Substitute ( ) ( ) exp( / 2)y H y y

Asymptotic solution in the limit of very large y:

2( ) exp( / 2)ny y y

Check:

Equation for H:


Simple Harmonic Oscillator (3)Must solve this ODE by the power-series method (Frobenius method); this is done as an example in 2246.

0

( ) pp

p

H y a y

We find:

•The series for H(y) must terminate in order to obtain a normalisable solution

•Can make this happen after n terms for either even or odd terms in series (but not both) by choosing

Label resulting functions of H by the values of n that we choose.

Hn is known as the nth Hermite polynomial.


The Hermite polynomials

For reference, first few Hermite polynomials are:

0

1

22

33

4 24

1;

2 ;

4 2;

8 12 ;

16 48 12.

H y

H y y

H y y

H y y y

H y y y

NOTE:

Hn contains yn as the highest power.

Each H is either an odd or an even function, according to whether n is even or odd.


Simple Harmonic Oscillator (4)Transforming back to the original variable x, the wavefunction becomes:

Probability per unit length of finding the particle is:

Compare classical result: probability of finding particle in a length δx is proportional to the time δt spent in that region:

For a classical particle with total energy E, velocity is given by


Notes

• “Zero-point energy”: • “Quanta” of energy:• Even and odd solutions• Applies to any simple harmonic oscillator, including

– Molecular vibrations– Vibrations in a solid (hence phonons)– Electromagnetic field modes (hence photons), even though this field

does not obey exactly the same Schrődinger equation

• You will do another, more elegant, solution method (no series or Hermite polynomials!) next year

• For high-energy states, probability density peaks at classical turning points (correspondence principle)


4 Postulates of QM

This section puts quantum mechanics onto a more formal mathematical footing by specifying those postulates of the theory which cannot be derived from classical physics.

Main ingredients:1. The wave function (to represent the state of the system);2. Hermitian operators (to represent observable quantities);3. A recipe for identifying the operator associated with a given

observable;4. A description of the measurement process, and for

predicting the distribution of outcomes of a measurement;5. A prescription for evolving the wavefunction in time (the

time-dependent Schrődinger equation)


4.1 The wave functionPostulate 4.1: There exists a wavefunction Ψ that is a continuous, square-integrable, single-valued function of the coordinates of all the particles and of time, and from which all possible predictions about the physical properties of the system can be obtained.

Examples of the meaning of “The coordinates of all the particles”:

For a single particle moving in one dimension:

For a single particle moving in three dimensions:

For two particles moving in three dimensions:

The modulus squared of Ψ for any value of the coordinates is the probability density (per unit length, or volume) that the system is found with that particular coordinate value (Born interpretation).


4.2 Observables and operators

1 1 2 2 1 1 2 2

1 2

1 2

[ ] [ ] [ ]

(for arbitrary functions and and

constants and )

L c f c f c L f c L f

f f

c c

1

2

3

4

ˆ [ ] 2

ˆ [ ]

ˆ [ ]

dˆ [ ]d

L f f

L f xf

L f f

fL f

x

Postulate 4.2.1: to each observable quantity is associated a linear, Hermitian operator (LHO).

An operator is linear if and only if

Examples: which of the operators defined by the following equations are linear?

Note: the operators involved may or may not be differential operators (i.e. may or may not involve differentiating the wavefunction).


Hermitian operators

*

* *ˆ ˆ( )d ( )d (4.2)f Og x g Of x

*

ij jiM M

ijM

An operator O is Hermitian if and only if:

for all functions f,g vanishing at infinity.

Compare the definition of a Hermitian matrix M:

Analogous if we identify a matrix element with an integral:

* ˆ( )di jf Of x

(see 3226 course for more detail…)


Hermitian operators: examplesThe operator is Hermitianx

dThe operator is not Hermitian.

dx

2

2

dThe operator is Hermitian.

dx


Eigenvectors and eigenfunctionsPostulate 4.2.2: the eigenvalues of the operator represent the possible results of carrying out a measurement of the corresponding quantity.

Definition of an eigenvalue for a general linear operator:

Example: the time-independent Schrődinger equation:

Compare definition of an eigenvalue of a matrix:


Postulate 4.2.3: immediately after making a measurement, the wavefunction is identical to an eigenfunction of the operator corresponding to the eigenvalue just obtained as the measurement result.

Proof:

Ensures that we get the same result if we immediately re-measure the same quantity.

Important fact: The eigenvalues of a Hermitian operator are real (like the eigenvalues of a Hermitian matrix).


4.3 Identifying the operators

ˆ xp ix

x x ˆ r r

Postulate 4.3: the operators representing the position and momentum of a particle are

Other operators may be obtained from the corresponding classical quantities by making these replacements.

Examples:

The Hamiltonian (representing the total energy as a function of the coordinates and momenta)

Angular momentum:

(one dimension)

ˆ i ix y z

p i j k

(three dimensions)


Eigenfunctions of momentumThe momentum operator is Hermitian, as required:

Its eigenfunctions are plane waves:


Orthogonality of eigenfunctions

ˆ ˆ; with n n n m m m n mQ q Q q q q

The eigenfunctions of a Hermitian operator belonging to different eigenvalues are orthogonal.

* d 0n m x

Proof:

If then


Orthonormality of eigenfunctionsWhat if two eigenfunctions have the same eigenvalue? (In this case the eigenvalue is said to be degenerate.)

Any linear combination of these eigenfunctions is also an eigenfunction with the same eigenvalue:

So we are free to choose as the eigenfunctions two linear combinations that are orthogonal.

If the eigenfunctions are all orthogonal and normalized, they are said to be orthonormal.


Orthonormality of eigenfunctions: example

Consider the solutions of the time-independent Schrődinger equation (energy eigenfunctions) for an infinite square well:

We chose the constants so that normalization is correct:


Complete sets of functionsThe eigenfunctions φn of a Hermitian operator form a complete set, meaning that any other function satisfying the same boundary conditions can be expanded as

( ) ( )n nn

x a x

If the eigenfunctions are chosen to be orthonormal, the coefficients an can be determined as follows:

We will see the significance of such expansions when we come to look at the measurement process.


Normalization and expansions in complete sets

The condition for normalizing the wavefunction is now

If the eigenfunctions φn are orthonormal, this becomes

Natural interpretation: the probability of finding the system in the state φn(x) (as opposed to any of the other eigenfunctions) is

2

na


Expansion in complete sets: example


4.4 Eigenfunctions and measurement

( ) ( ).n nn

x a x

Postulate 4.4: suppose a measurement of the quantity Q is made, and that the (normalized) wavefunction can be expanded in terms of the (normalized) eigenfunctions φn of the corresponding operator as

Then the probability of obtaining the corresponding eigenvalue qn as the measurement result is

2

na

What is the meaning of these “probabilities” in discussing the properties of a single system?

Still a matter for debate, but usual interpretation is that the probability of a particular result determines the frequency of occurrence of that result in measurements on an ensemble of similar systems.

Corollary: if a system is definitely in eigenstate φn, the result measuring Q is definitely the corresponding eigenvalue qn.


Commutators

ˆ ˆˆ ˆ (in general)QR RQ

In general operators do not commute: that is to say, the order in which we allow operators to act on functions matters:

For example, for position and momentum operators:

We define the commutator as the difference between the two orderings:

ˆ ˆ ˆˆ ˆ ˆ,Q R QR RQ

So, for position and momentum:

Two operators commute only if their commutator is zero.


Compatible operatorsTwo observables are compatible if their operators share the same eigenfunctions (but not necessarily the same eigenvalues).

Consequence: two compatible observables can have precisely-defined values simultaneously.

Compatible operators commute with one another:

Measure observable Q, obtain result qm (an eigenvalue of Q)

Wavefunction of system is corresponding eigenfunction φm

Measure observable R, definitely obtain result rm (the corresponding eigenvalue of R)

Wavefunction of system is still corresponding eigenfunction φm

Re-measure Q, definitely obtain result qm once again

Can also show the converse: any two commuting operators are compatible.

Consider a general wavefunction ( ) ( )n nn

x a x Expansion in terms of joint eigenfunctions of both operators


Example: measurement of position


Example: measurement of position (2)


Expectation valuesThe average (mean) value of measurements of the quantity Q is therefore the sum of the possible measurement results times the corresponding probabilities:

2.n n

n

Q a qWe can also write this as:


4.5 Evolution of the system

Postulate 4.5: Between measurements (i.e. when it is not disturbed by external influences) the wave-function evolves with time according to the time-dependent Schrődinger equation.

ˆ ,i Ht

Hamiltonian operator.

This is a linear, homogeneous differential equation, so the linear combination of any two solutions is also a solution: the superposition principle.


Calculating time dependence using expansion in energy eigenfunctions

( , ) exp( / ) ( )n nx t iE t x

ˆn n nH E

Suppose the Hamiltonian is time-independent. In that case we know that solutions of the time-dependent Schrődinger equation exist in the form:

where the wavefunctions ψ(x) and the energy E correspond to one solution of the time-independent Schrődinger equation:

We know that all the functions ψn together form a complete set, so we can expand

( ,0) ( )n nn

x a x Hence we can find the complete time dependence (superposition principle):


Time-dependent behaviour: exampleSuppose the state of a particle in an infinite square well at time t=0 is a `superposition’ of the n=1 and n=2 states

Wave function at a subsequent time t

Probability density


Rate of change of expectation valueConsider the rate of change of the expectation value of a quantity Q:

*d d ˆ( )dd d

QQ x

t t


Example 1: Conservation of probability

Rate of change of total probability that the particle may be found at any point:

Total probability conserved (related to existence of a well defined probability flux – see §3.4)

2dx

t

Total probability is the “expectation value” of the operator 1.


Example 2: Conservation of energy

Even although the energy of a system may be uncertain (in the sense that measurements of the energy made on many copies of the system may be give different results) the average energy is always conserved with time.

Consider the rate of change of the mean energy:

*

-

ˆ dE H xt t


5.1 Angular momentum operators

d d( )

d d

( ) ( ) 0.

t t

m

Lr p r p r ×p

pp r F

Angular momentum is a very important quantity in three-dimensional problems involving a central force (one that is always directed towards or away from a central point). In that case it is classically a conserved quantity:

We can write down a quantum-mechanical operator for it by applying our usual rules:

ˆ ˆ ˆ ( )i L = r ×p r

Individual components:

Central pointr

F

The origin of r is the same central point towards/away from which the force is directed.

Reading: Rae Chapter 5; B&J§§6.1,6.3; B&M§§6.2-6.5


5.2 Commutation relations***Remember:

[ , ]

[ , ]

[ , ]

x

y

z

x p i

y p i

z p i

The different components of angular momentum do not commute with one another.

[ , ]x yL L

By similar arguments get the cyclic permutations:


Commutation relations (2)

2 2 2 2ˆ ˆ ˆ ˆx y zL L L L

The different components of L do not commute with one another, but they do commute with the (squared) magnitude of the angular momentum vector:

Important consequence: we cannot find simultaneous eigenfunctions of all three components.

But we can find simultaneous eigenfunctions of one component (conventionally the z component) and L2

2 2 2,

, ,

A B A B ABA ABA BA

A A B A B A

Note a useful formula:


5.3 Angular momentum in spherical polar coordinates

1 1ˆ ˆ ˆ

sinr r r r

e e e

ˆi i rr L r e

zL i

Spherical polar coordinates are the natural coordinate system in which to describe angular momentum. In these coordinates,

On this slide, hats refer to unit vectors, not operators.

So the full (vector) angular momentum operator can be written

x

y

z

θ

φ

r

To find z-component, note that unit vector k in z-direction satisfies

(see 2246)

ˆrrr e


L2 in spherical polar coordinates2 2

2 2 2

( ) ( )

( )( )r

L L L r r

r r

22 2

2 2

1 1sin

sin sinL

Depends only on angular behaviour of wavefunction. Closely related to angular part of Laplacian (see 2246 and Section 6). 2

2 22 2 2

1 Lr

r r r r

On this slide, hats refer to unit vectors, not operators.


5.4 Eigenvalues and eigenfunctions Look for simultaneous eigenfunctions of L2 and one component of L (conventional to choose Lz)

Eigenvalues and eigenfunctions of Lz:

Physical boundary condition: wave-function must be single-valued

( 2 ) ( ) Quantization of angular momentum about z-axis (compare Bohr model)


Eigenvalues and eigenfunctions (2)

( , ) ( ) ( ) exp(i ) ( )f m Now look for eigenfunctions of L2, in the form

(ensures solutions remain eigenfunctions of Lz, as we want)

Eigenvalue condition becomes

2 2 2ˆ [exp(i ) ( )] exp(i ) ( ) Let eigenvalueL m m


The Legendre equationMake the substitution

cos

dsin

d

This is exactly the Legendre equation, solved in 2246 using the Frobenius method.


Legendre polynomials and associated Legendre functions

( 1), where 0,1,2,l l l

In order for solutions to exist that remain finite at μ=±1 (i.e. at θ=0 and θ=π) we require that the eigenvalue satisfies

The finite solutions are then the associated Legendre functions, which can be written in terms of the Legendre polynomials:

/ 22 d( ) (1 ) ( )

d

m

mml lP P

where m is an integer constrained to lie between –l and +l.

Legendre polynomials:

(like SHO, where we found restrictions on energy eigenvalue in order to produce normalizable solutions)


Spherical harmonics

00

11

01

11

1( , )

4

3 3 ( )( , ) sin exp( )

8 8

3 3( , ) cos

4 4

3 3 ( )( , ) sin exp( )

8 8

Y

x iyY i

r

zY

r

x iyY i

r

The full eigenfunctions can also be written as spherical harmonics:

Because they are eigenfunctions of Hermitian operators with different eigenvalues, they are automatically orthogonal when integrated over all angles (i.e. over the surface of the unit sphere). The constants C are conventionally defined so the spherical harmonics obey the following important normalization condition:

First few examples (see also 2246):

Remember

sin cos

sin sin

cos

x r

y r

z r


Shapes of the spherical harmonics

Imaginary

Real

z

x

y

0

0

1

4m

l

l

m

Y

1

1

3sin exp( )

8m

l

l

m

Y i

1

0

3cos

4m

l

l

m

Y

11Y

11Re[ ]Y

01Y0

0Y

To read plots: distance from origin corresponds to magnitude (modulus) of plotted quantity; colour corresponds to phase (argument).

(Images from http://odin.math.nau.edu/~jws/dpgraph/Yellm.html)

http://odin.math.nau.edu/~jws/dpgraph/dpgfiles/Yellm/Y00.dpg



http://odin.math.nau.edu/~jws/dpgraph/dpgfiles/Yellm/ReY11.dpg


Shapes of spherical harmonics (2)

(Images from http://odin.math.nau.edu/~jws/dpgraph/Yellm.html)

Imaginary

Real

z

x

y

2

2

0

5(3cos 1)

16m

l

l

m

Y

2

1

15sin cos exp(i )

8m

l

l

m

Y

2

2

2

15sin exp(2i )

32m

l

l

m

Y

12Y

12Re[ ]Y

22Y

22Re[ ]Y

02Y

To read plots: distance from origin corresponds to magnitude (modulus) of plotted quantity; colour corresponds to phase (argument).







5.5 The vector model for angular momentum***

2 2ˆEigenvalues of are ( 1) , with 0,1,2,L l l l

To summarize:

ˆEigenvalues of are , with , , 1,0,1,zL m m l l

These states do not correspond to well-defined values of Lx and Ly, since these operators do not commute with Lz.

l is known as the principal angular momentum quantum number: determines the magnitude of the angular momentum

m is known as the magnetic quantum number: determines the component of angular momentum along a chosen axis (the z-axis)

Semiclassical picture: each solution corresponds to a cone of angular momentum vectors, all with the same magnitude and the same z-component.


The vector model (2)

Example: l=2

Magnitude of angular momentum is

Component of angular momentum in z direction can be Lx

Ly

Lz

L


6.1 The three-dimensional square wellConsider a particle which is free to move in three dimensions everywhere within a cubic box, which extends from –a to +a in each direction. The particle is prevented from leaving the box by infinitely high potential barriers. x

yz

Time-independent Schrödinger equation within the box is free-particle like:

Separation of variables: takex, or y, or z( , , ) ( ) ( ) ( )x y z X x Y y Z z

V V

V(x)

-a a

0V

with boundary conditions ( ) ( ) ( ) 0.X a Y a Z a

Reading: Rae §3.2, B&J §7.4; B&M §5.11


Three-dimensional square well (2)

Substitute in Schrödinger equation:

Divide by XYZ:

Three effective one-dimensional Schrödinge equations.


Three-dimensional square well (3)Wavefunctions and energy eigenvalues known from solution to one-dimensional square well (see §3.2).

Total energy is


6.2 The Hamiltonian for a hydrogenic atom***

2 22

0

ˆ2 4e

ZeH

m r

; electron mass, nuclear masse Ne N

e N

m mm m

m m

For a hydrogenic atom or ion having nuclear charge +Ze and a single electron, the Hamiltonian is

The natural coordinate system to use is spherical polar coordinates. In this case the Laplacian operator becomes (see 2246):

r

+ZeNote: for greater accuracy we should use the reduced mass corresponding to the relative motion of the electron and the nucleus (since nucleus does not remain precisely fixed – see 1B2x):

-e

Note spherical symmetry – potential depends only on r

This means that the angular momentum about any axis, and also the total angular momentum, are conserved quantities: they commute with the Hamiltonian, and can have well-defined values in the energy eigenfunctions of the system.

2

2ˆd

ˆ ˆ[ , ] 0 0d

ˆdˆ ˆ[ , ] 0 0

d

z

z

LH L

t

LH L

t

Reading: Rae §§3.3-3.4, B&M Chapter 7, B&J §7.2 and §7.5


6.3 Separating the variablesWrite the time-independent Schrődinger equation as:

Now look for solutions in the form

( , , ) ( ) ( , )r R r Y

Substituting into the Schrődinger equation:


The angular equationWe recognise that the angular equation is simply the eigenvalue condition for the total angular momentum operator L2:

This means we already know the corresponding eigenvalues and eigenfunctions (see §5):

Note: all this would work for any spherically-symmetric potential V(r), not just for the Coulomb potential.


6.4 Solving the radial equation

Now the radial part of the Schrődinger equation becomes:

Define a new unknown function χ by: ( )( )

rR r

r

Note that this depends on l, but not on m: it therefore involves the magnitude of the angular momentum, but not its orientation.


The effective potentialThis corresponds to one-dimensional motion with the effective potential

2 2

eff 20

( 1)( )

4 2 e

Ze l lV r

r m r

First term:

Second term: r

V(r)


Atomic units***Atomic units: there are a lot of physical constants in these expressions. It makes atomic problems much more straightforward to adopt a system of units in which as many as possible of these constants are one. In atomic units we set:

2 1

23 2

0

20

2

Planck constant 1 (dimensions [ ])

Electron mass 1 (dimensions [ ])

Constant apearing in Coulomb's law 1 (dimensions [ ])4

It follows that:

4 Unit of length = 5.2917

e

e

ML T

m M

eML T

e m

110

2218

20

7 10 m=Bohr radius,

Unit of energy = 4.35974 10 J 27.21159eV=Hartree, 4

eh

a

meE

In this unit system, the radial equation becomes


Solution near the nucleus (small r)For small values of r the second derivative and centrifugal terms dominate over the others.

Try a solution to the differential equation in this limit as

We want a solution such that R(r) remains finite as r→0, so take


Asymptotic solution (large r)

( ) ( ) exp( )r F r r

Now consider the radial equation at very large distances from the nucleus, when both terms in the effective potential can be neglected. We are looking for bound states of the atom, where the electron does not have enough energy to escape to infinity:

Inspired by this, let us rewrite the solution in terms of yet another unknown function, F(r):

2

Put 2

E


Differential equation for FCan obtain a corresponding differential equation for F:

This equation is solved in 2246, using the Frobenius (power-series) method.

( ) k p p kp p

p p

F r r a r a r The indicial equation gives or 1k l l

1regular solution behaves like for small .lF r r


Properties of the series solution

where is an integer : 1, 2Z

n n l n l l

Hence the series must terminate after a finite number of terms. This happens only if

So the energy is

If the full series found in 2246 is allowed to continue up to an arbitrarily large number of terms, the overall solution behaves like

( ) exp(2 )

( ) exp(2 ) exp( ) exp( )

F r r

r r r r

Note that once we have chosen n, the energy is independent of both m (a feature of all spherically symmetric systems, and hence of all atoms) and l (a special feature of the Coulomb potential, and hence just of hydrogenic atoms).

n is known as the principal quantum number. It defines the “shell structure” of the atom.

(not normalizable)


6.5 The hydrogen energy spectrum and wavefunctions***

2

2 (in atomic units)

2n

ZE

n

, ( 1), 0, ( 1),m l l l l

12

0

(2 1) .n

l

l n

For each value of n=1,2,… we have a definite energy:

For each value of n, we can have n possible values of the total angular momentum quantum number l:

For each value of l and n we can have 2l+1 values of the magnetic quantum number m:

l=0,1,2,…,n-1

The total number of states (statistical weight) associated with a given energy En is therefore

l=0 l=1 l=2 l=3

Traditional nomenclature:l=0: s states (from “sharp” spectral lines)l=1: p states (“principal”)l=2: d states (“diffuse”)l=3: f states (“fine”)…and so on alphabetically (g,h,i… etc)

Each solution of the time-independent Schrődinger equation is defined by the three quantum numbers n,l,m

2

in units 2 h

ZE E

0

-1


The radial wavefunctions

3/ 2

10 00

3/ 2

210 0 0

3/ 2

200 0 0

3/ 2 2

320 0 0

3/

310

( ) 2 exp( / )

1( ) exp

2 23

( ) 2 1 exp2 2 2

4( ) exp

3 327 10

4 2( )

9 3

ZR r Zr a

a

Z Zr ZrR r

a a a

Z Zr ZrR r

a a a

Z Zr ZrR r

a a a

ZR r

a

2

0 0 0

3/ 2 2 2

30 20 0 0 0

1 exp6 3

2 2( ) 2 1 exp

3 3 27 3

Zr Zr Zr

a a a

Z Zr Z r ZrR r

a a a a

Radial wavefunctions Rnl depend on principal quantum number n and angular momentum quantum number l (but not on m)

Full wavefunctions are:

Normalization chosen so that:

Note:

Probability of finding electron between radius r and r+dr is

Only s states (l=0) are finite at the origin.

Radial functions have (n-l-1) zeros.


Comparison with Bohr model***

, 1, 2,3,zL n n

20

0, Bohr radiusn

n ar a

Z

Angular momentum (about any axis) assumed to be quantized in units of Planck’s constant:

Electron otherwise moves according to classical mechanics and has a single well-defined orbit with radius

Energy quantized and determined solely by angular momentum:

Bohr model Quantum mechanics

2

2, Hartree

2n h h

ZE E E

n

, , ,zL m m l l

020

1 , Bohr radiusZ

ar n a

2

2, Hartree

2n h h

ZE E E

n

Angular momentum (about any axis) shown to be quantized in units of Planck’s constant:

Energy quantized, but is determined solely by principal quantum number, not by angular momentum:

Electron wavefunction spread over all radii. Can show that the quantum mechanical expectation value of the quantity 1/r satisfies


6.6 The remaining approximations

• This is still not an exact treatment of a real H atom, because we have made several approximations.– We have neglected the motion of the nucleus. To fix this we would

need to replace me by the reduced mass μ (see slide 1).– We have used a non-relativistic treatment of the electron and in

particular have neglected its spin (see §7). Including these effects gives rise to

• “fine structure” (from the interaction of the electron’s orbital motion with its spin), and

• “hyperfine structure” (from the interaction of the electron’s spin with the spin of the nucleus)

– We have neglected the fact that the electromagnetic field acting between the nucleus and the electron is itself a quantum object. This leads to “quantum electrodynamic” corrections, and in particular to a small “Lamb shift” of the energy levels.


7.1 Atoms in magnetic fields

Interaction of classically orbiting electron with magnetic field:

v

Orbit behaves like a current loop:

2

Loop current= (- sign because charge )2

Magnetic moment current area

=2 2

where (the Bohr magneton).2

e Be

Be

eve

r

ev e Lr m vr

r m

e

m

In the presence of a magnetic field B, classical interaction energy is:

r

Corresponding quantum mechanical expression (to a good approximation) involves the angular momentum operator:

μ

Reading: Rae Chapter 6; B&J §6.8, B&M Chapter 8 (all go further than 2B22)


Splitting of atomic energy levelsSuppose field is in the z direction. The Hamiltonian operator is

0ˆ ˆ ˆB z

z

BH H L

We chose energy eigenfunctions of the original atom that are eigenfunctions of Lz so these same states are also eigenfunctions of the new H.

0 0ˆ ;

ˆ .

m m

z m m

H E

L m


Splitting of atomic energy levels (2)

0B

Predictions: should always get an odd number of levels. An s state (such as the ground state of hydrogen, n=1, l=0, m=0) should not be split.

(2l+1) states with same energy: m=-l,…+l

(Hence the name “magnetic quantum number” for m.)

0B


7.2 The Stern-Gerlach experiment***

( ) F μ B

Study deflection of atoms in inhomogeneous magnetic field. Force on atoms is

Gerlach

N

S

Produce a beam of atoms with a single electron in an s state (e.g. hydrogen, sodium)

Results show two groups of atoms, deflected in opposite directions, with magnetic moments

B

Consistent neither with classical physics (which would predict a continuous distribution of μ) nor with our quantum mechanics so far (which always predicts an odd number of groups, and just one for an s state).

http://images.google.co.uk/imgres?imgurl=www.th.physik.uni-frankfurt.de/~jr/gif/phys/gerlach.jpg&imgrefurl=http://www.th.physik.uni-frankfurt.de/~jr/physlist.html&h=600&w=435&prev=/images%3Fq%3Dgerlach%26svnum%3D10%26hl%3Den%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8


7.3 The concept of spin***Try to understand these results by analogy with what we know about the ordinary (“orbital”) angular momentum: must be due to some additional source of angular momentum that does not require motion of the electron. Known as “spin”.

Introduce new operators to represent spin, assumed to have same commutation relations as ordinary angular momentum:

Goudsmit Uhlenbeck

Corresponding eigenfunctions and eigenvalues:

Pauli

(will see in Y3 that these equations can be derived directly from the commutation relations)

http://images.google.co.uk/imgres?imgurl=w3.hep.uiuc.edu/~leigh/class/spin/mugs/Uhlenbeck.jpeg&imgrefurl=http://w3.hep.uiuc.edu/~leigh/class/spin/mugs/mugs.html&h=232&w=190&prev=/images%3Fq%3Duhlenbeck%26svnum%3D10%26hl%3Den%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8

http://images.google.co.uk/imgres?imgurl=www.childrenofthemanhattanproject.org/HF/Photos%2520-%2520Men/goudsmit.jpg&imgrefurl=http://www.childrenofthemanhattanproject.org/HISTORY/H-06f.htm&h=851&w=540&prev=/images%3Fq%3Dgoudsmit%26svnum%3D10%26hl%3Den%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8%26sa%3DN


Spin quantum numbers for an electron

0ˆˆ ˆ ˆ( )

2 (Dirac's relativistic theory)(beyond 2222)

2.00231930437 (Quantum Electrodynamics)

BH H g

g

g

B L S

General interaction with magnetic field:

From the Stern-Gerlach experiment, we know that electron spin along a given axis has two possible values.

So, choose

But we also know from Stern-Gerlach that magnetic moments associated with the two possibilities are B

So, have

Spin angular momentum is twice as “effective” at producing magnetic moment as orbital angular momentum.


A complete set of quantum numbers

Hence the complete set of quantum numbers for the electron in the H atom is: n,l,m,s,ms.

Corresponding to a full wavefunction

Note that the spin functions χ do not depend on the electron coordinates r,θ,φ; they represent a purely internal degree of freedom.

H atom in magnetic field, with spin included:


7.4 Combining different angular momenta

1 12 2,

, ,j

j l l

m j j

So, an electron in an atom has two sources of angular momentum:

•Orbital angular momentum (arising from its motion through the atom)

•Spin angular momentum (an internal property of its own).

To think about the total angular momentum produced by combining the two, use the vector model once again:

Lx

Ly

Lz

Vector addition between orbital angular momentum L (of magnitude L) and spin S (of magnitude S): produces a resulting angular momentum vector J: quantum mechanics says its magnitude lies somewhere between |L-S| and L+S.(in integer steps).

L

S

L+S

|L-S|

For a single electron, corresponding `total angular momentum’ quantum numbers are

Determines length of resultant angular momentum vectorDetermines orientation

J L S

L

S


Example: the 1s and 2p states of hydrogen

The 1s state:

The 2p state:


Combining angular momenta (2)The same rules apply to combining other angular momenta, from whatever source.

For example for two electrons in an excited state of He atom, one in 1s state and one in 2p state (defines what is called the 1s2p configuration in atomic spectroscopy):

First construct combined orbital angular momentum L of both electrons:

Then construct combined spin S of both electrons:

Hence there are two possible terms (combinations of L and S):

…and four levels (possible ways of combining L and S to get different total angular momentum quantum numbers)

1 11 1 2 22 20; ; 1;l s l s


Term notationSpectroscopists use a special notation to describe terms and levels:

2 1SJL

•The first (upper) symbol is a number giving the number of spin states corresponding to the total spin S of the electrons

•The second (main) symbol is a letter encoding the total orbital angular momentum L of the electrons:

•S denotes L=0•P denotes L=1•D denotes L=2 (and so on);

•The final (lower) symbol gives the total angular momentum J obtained from combining the two.

Example: terms and levels from previous page would be:


7.5 Wavepackets and the Uncertainty Principle revisited (belongs in §4 – non-examinable)

2( )x

Can think of the Uncertainty Principle as arising from the structure of wavepackets. Consider a normalized wavefunction for a particle located somewhere near (but not exactly at) position x0

Can also write this as a Fourier transform (see 2246):

2( )k

Probability density:

x

k

(expansion in eigenstates of momentum)


Fourier transform of a Gaussian

21/ 42 0

2

21/ 42 0

0 02

2 21/ 42 2 20

0 2

1( ) ( ) exp( )d

2

( )12 exp exp( )d

42

( )12 exp( i ) exp ( ) d

42

[( ) 2 ]12 exp( i ) exp d

42

1exp

2

k x ikx x

x xikx x

x xkx ik x x x

x x ikkx k x

1/ 42 2 2 2 2 20 0

1/ 42 2 20

1/ 42 2

02

1/ 22 2 2

2

2

2 2

( i ) exp( ) 2 exp( ' / 4 )d ' with ' 2i ,d ' d

1exp( i ) exp( ) 2 2

2

exp( i ) exp( )2

( ) exp( 2 )2

( ) d2 2

kx k x x x x x k x x

kx k

kx k

k k

k k

1/ 2

1

3

2 2 2 2 2exp( / )d ; exp( / )d2

ax a x a x x a x


Wavepackets and Uncertainty Principle (2)

In fact, can show that this form of wavepacket (“Gaussian wavepacket”) minimizes the product of Δx and Δp, so:

Mean-squared uncertainty in postion

Mean-squared uncertainty in momentum:

Mean momentum:

2x p


Wavepackets and Uncertainty Principle (3)

SummaryThree ways of thinking of Uncertainty principle:(1) Arising from the physics of the interaction of different types of measurement apparatus

with the system (e.g. in the gamma-ray microscope);(2) Arising from the properties of Fourier transforms (narrower wavepackets need a wider

range of wavenumbers in their Fourier transforms);(3) Arising from the fact that x and p are not compatible quantities (do not commute), so

they cannot simultaneously have precisely defined values.

General result (see third year, or Rae §4.5):

12

ˆ ˆFor general non-commuting operators ,

ˆ ˆ,

ˆ ˆwhere , are RMS uncertainties in , .

Q R

q r Q R

q r Q R